Chapter 7, Problem 13P

In the circuit of Fig. 7.93,

$\begin{array}{l}v(t)=80e^{-{10}^{3}t}\text{V},t>0\\ i(t)=5e^{-{10}^{3}t}\text{mA},t>0\end{array}$

1. (a) Find R, L, and τ.
2. (b) Calculate the energy dissipated in the resistance for 0 < t < 0.5 ms.

Figure 7.93

For Prob. 7.13.

To determine

Find the value of resistance R, inductance L and also find the time constant $\tau$ in the circuit.

Answer to Problem 13P

The value of resistance R in the circuit is $16\text{kΩ}$, the value of inductance L in the circuit is $16\text{H}$ and the time constant in the circuit is $1\text{ms}$.

Explanation of Solution

Given data:

The voltage $\left(v\left(t\right)\right)$ is $80e^{-{10}^{3}t}\text{V}$.

The current $\left(i\left(t\right)\right)$ is $5e^{-{10}^{3}t}\text{mA}$.

Formula used:

Write the general expression to find the voltage across the resistor using Ohms law.

$v\left(t\right)=i\left(t\right)R$ (1)

Here,

$i\left(t\right)$ is the current response in the circuit, and

$R$ is the resistance in the circuit.

Write the expression to find the time constant for RL circuit.

$\tau =\frac{L}{R}$ (2)

Here,

R is the resistance of the resistor, and

L is the inductance of the inductor.

Write the general expression to find the voltage response in the RL Circuit.

$v\left(t\right)=V_0{e}^{-\frac{t}{\tau }}$ (3)

Here,

$\tau$ is the time constant for the RC circuit, and

$V_0$ is the initial voltage at the capacitor terminal.

Calculation:

Substitute $80e^{-{10}^{3}t}\text{V}$ for $v\left(t\right)$ and $5e^{-{10}^{3}t}\text{mA}$ for $i\left(t\right)$ in equation (1) to find the resistance R.

$80e^{-{10}^{3}t}\text{V}=\left(5e^{-{10}^{3}t}\text{mA}\right)R$

Rearrange the above equation to find resistance R.

$\begin{array}{c}R=\frac{80e^{-{10}^{3}t}\text{V}}{5e^{-{10}^{3}t}\text{mA}}\\ =\frac{80e^{-{10}^{3}t}\text{V}}{5e^{-{10}^{3}t}\times {10}^{-3}\text{A}}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =16\times {10}^3\frac{\text{V}}{\text{A}}\\ =16\text{k}\Omega \left\{\because 1\text{k}={10}^3,1\Omega =\frac{1\text{V}}{1\text{A}}\right\}\end{array}$

Compare the given voltage $v\left(t\right)=80e^{-{10}^{3}t}\text{V}$ and equation (3) for the time constant $\tau$.

$\frac{1}{\tau }=\frac{{10}^3}{1\text{s}}$ (4)

Rearrange the equation (4) to find the time constant $\tau$.

$\begin{array}{c}\tau =\frac{1}{{10}^3}\text{s}\\ ={10}^{-3}\text{s}\\ =1\text{ms}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Substitute $\frac{1}{{10}^3}\text{s}$ for $\tau$ and $16\text{k}\Omega$ for R in equation (2).

$\frac{1}{{10}^3}\text{s}=\frac{L}{16\text{k}\Omega }$ (5)

Rearrange the equation (5) to find the inductance L in Henry.

$\begin{array}{c}L=\left(\frac{1}{{10}^3}\text{s}\right)\left(16\text{k}\Omega \right)\\ =\left(\frac{1}{{10}^3}\text{s}\right)\left(16\times {10}^3\Omega \right)\left\{\because 1\text{k}={10}^3\right\}\\ =16\Omega \cdot \text{s}\\ =16\text{H}\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\end{array}$

Conclusion:

Thus, the value of resistance R in the circuit is $16\text{kΩ}$, the value of inductance L in the circuit is $16\text{H}$ and the time constant in the circuit is $1\text{ms}$.

To determine

Find the value of energy dissipated in the resistance for $0<t<0.5\text{ms}$.

Answer to Problem 13P

The value of energy dissipated in the resistance for $0<t<0.5\text{ms}$. is $126.42\mu \text{J}$.

Explanation of Solution

Given data:

The energy dissipated in the resistance for the time limits $0<t<0.5\text{ms}$.

Formula used:

Write the expression to find the energy dissipated in the resistor.

$w={\displaystyle \underset{0}{\overset{t}{\int }}pdt}$ (6)

Here,

$p$ is the power in the circuit.

Write the expression to find the power dissipated in the resistor.

$p=v\left(t\right)i\left(t\right)$ (7)

Here,

$v\left(t\right)$ is the voltage in the circuit, and

$i\left(t\right)$ is the current in the circuit.

Substitute equation (7) in equation (6) to find the energy dissipated in the resistor w.

$w={\displaystyle \underset{0}{\overset{t}{\int }}v\left(t\right)i\left(t\right)dt}$ (8)

Calculation:

Substitute $80e^{-{10}^{3}t}\text{V}$ for $v\left(t\right)$, $0.5\text{ms}$ for t and $5e^{-{10}^{3}t}\text{mA}$ for $i\left(t\right)$ in equation (8) to find the energy dissipated in the resistor w.

$\begin{array}{c}w={\displaystyle \underset{0}{\overset{0.5\text{ms}}{\int }}\left(80e^{-{10}^{3}t}\text{V}\right)\left(5e^{-{10}^{3}t}\text{mA}\right)dt}\\ ={\displaystyle \underset{0}{\overset{0.5\times {10}^{-3}\text{s}}{\int }}\left(80e^{-{10}^{3}t}\text{V}\right)\left(5e^{-{10}^{3}t}\times {10}^{-3}\text{A}\right)dt}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =0.4{\displaystyle \underset{0}{\overset{0.5\times {10}^{-3}\text{s}}{\int }}\left(e^{-{10}^{3}t}\text{V}\right)\left(e^{-{10}^{3}t}\text{A}\right)dt}\end{array}$$w=\left[0.4{\displaystyle \underset{0}{\overset{0.5\times {10}^{-3}\text{s}}{\int }}{e}^{-{10}^{3}t-{10}^{3}t}dt}\right]\text{V}\cdot \text{A}$ (9)

Rewrite the equation (9) as follows,

$\begin{array}{c}w=\left[0.4{\displaystyle \underset{0}{\overset{0.5\times {10}^{-3}\text{s}}{\int }}{e}^{-2000t}dt}\right]\text{W}\left\{\because 1\text{W}=1\text{V}.1\text{A}\right\}\\ =\left[0.4{\left(\frac{e^{-2000t}}{-2000}\right)}_0^{0.5\times {10}^{-3}\text{s}}\right]\text{W}\cdot \text{s}\\ =\left[-2\times {10}^{-4}{\left(e^{-2000t}\right)}_0^{0.5\times {10}^{-3}\text{s}}\right]\text{J}\left\{\because 1\text{J}=1\text{W}.1\text{s}\right\}\end{array}$

$w=\left[-2\times {10}^{-4}\left(e^{-2000\left(0.5\times {10}^{-3}\text{s}\right)}-e^{-2000\left(0\right)}\right)\right]\text{J}$ (10)

Reduce the equation (10) to find the energy dissipated in the resistor w in joules.

$\begin{array}{c}w=\left[-2\times {10}^{-4}\left(0.36788-1\right)\right]\text{J}\\ =\left[-2\times {10}^{-4}\left(-0.63212\right)\right]\text{J}\\ =\text{1}\text{.2642}\times {10}^{-4}\text{J}\end{array}$

Converting the unit J to $\mu \text{J}$.

$\begin{array}{c}w=\text{126}\text{.42}\times {10}^{-2}\times {10}^{-4}\text{J}\\ =\text{126}\text{.42}\times {10}^{-6}\text{J}\\ =\text{126}\text{.42}\mu \text{J}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the value of energy dissipated in the resistance for $0<t<0.5\text{ms}$. is $126.42\mu \text{J}$.